A Decision Process For Standard Antiderivatives

Why Use an Ordered Decision Process?

An integral can contain several recognizable features at once: a product, a rational expression, a composite function, or a trigonometric expression. Starting with the first method that comes to mind can create unnecessary algebra or lead to a method that does not fit. A short, ordered inspection helps you choose a reasonable first attempt.

The process is a guide, not a theorem that identifies one uniquely correct method for every integral. Different approaches can work, and an integral may require more than one method. The goal is to spot simple structure before committing to a longer calculation.

Definition: An antiderivative of \(f\) on an interval is a function \(F\) such that \(F'(x)=f(x)\) throughout that interval. An indefinite integral, \(\int f(x)\,dx\), represents the family of antiderivatives, written \(F(x)+C\), where \(C\) is an arbitrary constant.

The Decision Process

Simplify before choosing a method.
Factor, cancel only common factors, expand only when it helps, and consider polynomial division for a rational expression whose numerator degree is at least its denominator degree.

Check for a direct antiderivative or a simple rewrite.
Look for basic powers, exponentials, and trigonometric functions, and rewrite sums or quotients into simpler terms when appropriate.

Look for a composite function paired with its inner derivative.
If the integrand has the form \(f(g(x))g'(x)\), substitution is a natural candidate.

Classify products and trigonometric expressions.
A product may suit integration by parts. A product of trigonometric powers may suit trigonometric identities and substitution; a radical involving a quadratic expression may suit trigonometric substitution.

Inspect rational functions.
After any needed division, factor the denominator. Partial fractions may help when the rational function is proper and its denominator factors into manageable real linear or quadratic factors.

Compare the likely work.
Choose a method that simplifies the integrand and leaves a manageable integral. If a first attempt makes the expression more complicated, reconsider the classification.

The order matters because it tests inexpensive options first. For example, a rational expression might look like a partial-fractions problem, but division or cancellation could turn it into a sum of terms that are integrated directly. Likewise, a product is not automatically a reason to use integration by parts: one factor may be the derivative of an inner function in the other factor, making substitution simpler.

Substitution: Match the Inner Function and Its Derivative

Substitution is suggested when part of the integrand is a composite function and another factor supplies the derivative of its inner function, up to a constant multiple. The chain rule explains why this pattern works: differentiating \(F(g(x))\) produces \(F'(g(x))g'(x)\). Thus, if \(F'=f\), then \(F(g(x))\) is an antiderivative of \(f(g(x))g'(x)\).

\frac{d}{dx}F(g(x))=F'(g(x))g'(x)=f(g(x))g'(x).

This is the reverse-chain-rule basis for substitution. A proposed substitution must account for the differential as well as the inner function; the match can include a constant factor, but that factor must be retained.

Worked Example: A Denominator and Its Derivative

Evaluate \(\int \frac{2x}{x^2+5}\,dx\). The denominator contains \(x^2+5\), whose derivative is \(2x\), exactly the numerator. Set \(u=x^2+5\), so \(du=2x\,dx\). Then

\int \frac{2x}{x^2+5}\,dx =\int \frac{1}{u}\,du =\ln|u|+C =\ln(x^2+5)+C.

The absolute value can be omitted in the last expression because \(x^2+5>0\) for every real \(x\). To check, differentiate the result:

\frac{d}{dx}\ln(x^2+5) =\frac{2x}{x^2+5},

which is the original integrand.

Products and Trigonometric Expressions

If an integrand is a product and substitution does not give a useful match, consider integration by parts. The formula is

\int u\,dv=uv-\int v\,du.

Here \(u\) and \(v\) are differentiable functions, and \(dv\) is chosen so that it can be integrated to obtain \(v\). This method is useful when differentiating one factor simplifies it and integrating the other factor is manageable. Merely seeing a product is not enough: the resulting integral should be simpler than the original one.

Worked Example: A Polynomial–Trigonometric Product

Evaluate \(\int x\cos x\,dx\). Choose \(u=x\) and \(dv=\cos x\,dx\). Then \(du=dx\) and \(v=\sin x\). Applying integration by parts gives

\begin{aligned} \int x\cos x\,dx &=x\sin x-\int \sin x\,dx\\ &=x\sin x+\cos x+C. \end{aligned}

Differentiate to check:

\frac{d}{dx}\bigl(x\sin x+\cos x\bigr) =\sin x+x\cos x-\sin x =x\cos x.

The result differentiates to the original integrand.

Trigonometric expressions require a further distinction. Trigonometric integration uses identities and substitutions to integrate powers or products of trigonometric functions. Trigonometric substitution changes a variable to a trigonometric function, often to simplify a radical involving a quadratic expression. These are related tools, but they address different visible structures.

Worked Example: An Odd Power of Sine

Evaluate \(\int \sin^3 x\cos^2 x\,dx\). Because the sine power is odd, write \(\sin^3 x=\sin^2 x\sin x=(1-\cos^2 x)\sin x\). Set \(u=\cos x\), so \(du=-\sin x\,dx\). Then

\begin{aligned} \int \sin^3 x\cos^2 x\,dx &=\int (1-\cos^2 x)\cos^2 x\sin x\,dx\\ &=-\int (1-u^2)u^2\,du\\ &=-\int (u^2-u^4)\,du\\ &=-\frac{u^3}{3}+\frac{u^5}{5}+C\\ &=-\frac{\cos^3 x}{3}+\frac{\cos^5 x}{5}+C. \end{aligned}

For the check, differentiate the final expression:

\sin x\cos^2 x-\sin x\cos^4 x =\sin x\cos^2 x(1-\cos^2 x) =\sin^3 x\cos^2 x.

This recovers the integrand.

Worked Example: A Quadratic Radical

Consider \(\int \frac{1}{\sqrt{9-x^2}}\,dx\) for \(-3<x<3\). The expression \(9-x^2\) suggests the substitution \(x=3\sin\theta\). Choose \(-\frac{\pi}{2}<\theta<\frac{\pi}{2}\), where \(\cos\theta>0\). Then \(dx=3\cos\theta\,d\theta\), and

\sqrt{9-x^2} =\sqrt{9-9\sin^2\theta} =3\cos\theta.

Therefore,

\begin{aligned} \int \frac{1}{\sqrt{9-x^2}}\,dx &=\int \frac{3\cos\theta}{3\cos\theta}\,d\theta\\ &=\int 1\,d\theta\\ &=\theta+C\\ &=\arcsin\left(\frac{x}{3}\right)+C. \end{aligned}

To check the answer on \(-3<x<3\), differentiate:

\frac{d}{dx}\arcsin\left(\frac{x}{3}\right) =\frac{1/3}{\sqrt{1-x^2/9}} =\frac{1}{\sqrt{9-x^2}}.

The derivative is the original integrand.

Rational Functions and Partial Fractions

A rational function is a quotient of polynomials. Before applying partial fractions, check whether the fraction is proper: its numerator degree must be less than its denominator degree. If it is not proper, polynomial division comes first. For a proper rational function, factor the denominator as far as possible over the real numbers. Partial fractions are a candidate when this produces simpler terms whose antiderivatives are known.

Worked Example: Two Linear Denominator Factors

Evaluate \(\int \frac{3x+5}{(x+1)(x+2)}\,dx\). The fraction is proper, and the denominator has distinct linear factors. Seek constants \(A\) and \(B\) such that

\frac{3x+5}{(x+1)(x+2)} =\frac{A}{x+1}+\frac{B}{x+2}.

Multiplying by \((x+1)(x+2)\) gives \(3x+5=A(x+2)+B(x+1)\). Equating coefficients yields \(A+B=3\) and \(2A+B=5\). Subtracting the first equation from the second gives \(A=2\), and then \(B=1\). Thus

\begin{aligned} \int \frac{3x+5}{(x+1)(x+2)}\,dx &=\int\left(\frac{2}{x+1}+\frac{1}{x+2}\right)\,dx\\ &=2\ln|x+1|+\ln|x+2|+C. \end{aligned}

Differentiate to check:

\frac{2}{x+1}+\frac{1}{x+2} =\frac{2(x+2)+(x+1)}{(x+1)(x+2)} =\frac{3x+5}{(x+1)(x+2)}.

The derivative equals the original rational function wherever it is defined.

A Common Misclassification

A common mistake is to apply partial fractions before checking whether the rational function is proper. For example, it is incorrect to treat \(\frac{x^2+1}{x}\) as a constant multiple of \(\frac{1}{x}\): that would require \(x^2+1=A\) for a constant \(A\), which is impossible. First divide each numerator term by \(x\):

\frac{x^2+1}{x}=x+\frac{1}{x}.

Now integrate the simpler terms:

\int \frac{x^2+1}{x}\,dx =\int x\,dx+\int \frac{1}{x}\,dx =\frac{x^2}{2}+\ln|x|+C, \qquad x\ne 0.

Differentiating gives \(x+\frac{1}{x}=\frac{x^2+1}{x}\), confirming the correction. The practical lesson is to simplify and check the degree condition before choosing partial fractions.

Using the Process Well

The decision process narrows the choices; it does not replace judgment. An integral may combine structures, and one method may prepare the expression for another. Prefer the candidate that creates the clearest next step, rather than choosing a method solely because a familiar pattern appears somewhere in the integrand.

Takeaway: Simplify first, test direct patterns and reverse-chain-rule matches, then classify products, trigonometric forms, and rational functions. Choose the method whose conditions fit and whose next integral looks manageable.

Check Your Understanding

For each question, explain which feature of the integrand guides the choice.

What should you check about the degrees of the numerator and denominator before using partial fractions?
For \(\int 6x(x^2+4)^3\,dx\), what substitution is suggested, and what factor in the integrand supports it?
What is the difference between trigonometric integration and trigonometric substitution?
Why does seeing a product alone not guarantee that integration by parts is the best method?
What algebraic rewrite should come before integrating \(\frac{x^2+1}{x}\)?